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RC in Parallel

P

Peter

Jan 1, 1970
0
I'm trying to figure something out and so far no one has come up with an
answer.

If I have a voltage divider network (Just 2 resistors) and I bypass the
bottom one with a cap with an AC source in. How do I figure out the
impedance?

My goal is the following: put a sine wave in with say 4v p-p in @1kHz, have
an R1 and R2 (equal value) with a bypass cap on R2 (the other side of R2 is
connected to ground) and say I get 1v p-p out (the middle of R1 and R2 is my
output). I want to take that circuit and replace it with just R1 and R2
(keeping R1 the same value) and get the same amplitude out. I don't care
about phase right now.

Anyone have any ideas?
 
R

Roy Q.T.

Jan 1, 1970
0
I've had no need to figure anything remotely close to this in years but;
Don't you have to calculate XC or capacity reactance to get Z impedance
in an RC network ?

good luck !

Roy ~ E.E.Technician
 
Peter said:
I'm trying to figure something out and so far no one has come up with an
answer.

If I have a voltage divider network (Just 2 resistors) and I bypass the
bottom one with a cap with an AC source in. How do I figure out the
impedance?

My goal is the following: put a sine wave in with say 4v p-p in @1kHz, have
an R1 and R2 (equal value) with a bypass cap on R2 (the other side of R2 is
connected to ground) and say I get 1v p-p out (the middle of R1 and R2 is my
output). I want to take that circuit and replace it with just R1 and R2
(keeping R1 the same value) and get the same amplitude out. I don't care
about phase right now.

Anyone have any ideas?
If you already know what you want for Vout (1 volt)
and you know that Vin is 4 volts, why do you care
what the cap is doing, since it won't be in the
new circuit? You know its presence in the existing
circuit with equal Rs causes the the thing to
divide by 4.

Anyway: Vout = Vin*NewR2/(R1+NewR2) where Vout appears
across the NewR2

If you want to know the impedance of your existing
R2 and cap, solve the above equation for the new
R2. Then solve for Zc with R2new = R1*Zc/(R1+Zc).
Or figure the capacitave reactance at 1 khZ
Xc = 1/2*pi*f*C. In this case, Xc and Zc are
the same.

Ed
 
P

Palindr☻me

Jan 1, 1970
0
Peter said:
I'm trying to figure something out and so far no one has come up with an
answer.

If I have a voltage divider network (Just 2 resistors) and I bypass the
bottom one with a cap with an AC source in. How do I figure out the
impedance?

Do you mean the impedance as seen by the supply? In which
case, simply work out the impedance of one resistor in
series with a parallel combination of the capacitor, the
other resistor (and what ever load you might want to work it
out for).

Do you mean the impedance as seen by a load? In which case
it is the impedance of the parallel combination of the
resistor and capacitor, in series with the resistor (and the
impedance of ths source, if not considered negligable).
My goal is the following: put a sine wave in with say 4v p-p in @1kHz, have
an R1 and R2 (equal value) with a bypass cap on R2 (the other side of R2 is
connected to ground) and say I get 1v p-p out (the middle of R1 and R2 is my
output). I want to take that circuit and replace it with just R1 and R2
(keeping R1 the same value) and get the same amplitude out. I don't care
about phase right now.

The voltage source will see R1 in series with the parallel
combination of R2b and your load impedance. Assuming that
your load impedance is high enough to be insignificant, this
reduces to R1 in series with R2b. If you can't work out the
value of R2b - bearing in mind that the same current flows
through R1 and R2b but that the voltage across R1 is three
times the voltage across R2b, then you want to take up
needlework and/or cookery and leave electrickery to us girls.
 
B

Bob Eldred

Jan 1, 1970
0
Peter said:
I'm trying to figure something out and so far no one has come up with an
answer.

If I have a voltage divider network (Just 2 resistors) and I bypass the
bottom one with a cap with an AC source in. How do I figure out the
impedance?

My goal is the following: put a sine wave in with say 4v p-p in @1kHz, have
an R1 and R2 (equal value) with a bypass cap on R2 (the other side of R2 is
connected to ground) and say I get 1v p-p out (the middle of R1 and R2 is my
output). I want to take that circuit and replace it with just R1 and R2
(keeping R1 the same value) and get the same amplitude out. I don't care
about phase right now.

Anyone have any ideas?


Why do you have the resistor bypassed if you are trying to make a voltage
divider? In a high frequency attenuator, both resistors are in parallel with
capacitors and have their time constants equal like in a scope. But, I don't
understand what you are trying to do.
Bob
 
M

meqshesher

Jan 1, 1970
0
Peter said:
I'm trying to figure something out and so far no one has come up with an
answer.

If I have a voltage divider network (Just 2 resistors) and I bypass the
bottom one with a cap with an AC source in. How do I figure out the
impedance?

presented to the previous stage or the next stage?

"Bypass" means AC short, so the impeadance presented to the next stage will
be low (looking mostly at the cap impeadance)

My goal is the following: put a sine wave in with say 4v p-p in @1kHz, have
an R1 and R2 (equal value)

You have 1/2 voltage out R2/(R1+R2) (no cap)
with a bypass cap on R2 (the other side of R2 is
connected to ground) and say I get 1v p-p out (the middle of R1 and R2 is my
output).

Now your divider is 1/4 so Z of the cap is about 1/3 of R (estimate)
I want to take that circuit and replace it with just R1 and R2
(keeping R1 the same value) and get the same amplitude out.

So take the cap out? and have a voltage division of 1/4?

1/4 = R2/(R1+R2) so 4*R2=R1+R2 or R1= 3*R2

So if R1 is 1k then R2= 330 ohms and presented Z to previous stage is
about 1.3k and Z presented to next stage is about 250 or so. (need Z's of
the stages to have real answers)
 
P

Peter

Jan 1, 1970
0
The reason I'm trying this circuit is because I was trying to understand
bypass caps in a common emitter transistor circuit. Everytime I tried
figuring out the impedance looking into the base, I didn't get what I
expected.

So I built this circuit to learn from it then use what I learn and apply
that back into a common emitter circuit.
 
J

Jimmie

Jan 1, 1970
0
Peter said:
The reason I'm trying this circuit is because I was trying to understand
bypass caps in a common emitter transistor circuit. Everytime I tried
figuring out the impedance looking into the base, I didn't get what I
expected.

So I built this circuit to learn from it then use what I learn and apply
that back into a common emitter circuit.
How far off are you calculation? Remember to include the base /emitter
junction resistance. Mostly this value is insignificant unless the the total
emitter resistance is by passed. If the emitter resistor is completely
bipassed this may have the most significant effect on the input impedance
depending on the circuit design.
 
P

Peter

Jan 1, 1970
0
Well that's why I'm using THIS circuit so I don't have to worry about all
the factors of the transistor.

If I can figure out how to go from a 2 resistor series circuit with the
bottom resistor bypassed by a cap and the middle of the two resistors as my
Vout to just a 2 resistor series circuit with ONLY changing the bottom value
resistor to the equivilent of the R and C combiation and get the SAME Vout,
then I will know for sure that my emitter is all set and any variations I
get in calculations will be due to temp or whatever.

This is why this circuit is important. I don't care about phase. I ONLY care
about how to go from 1kHz 4Vp-p with both resistors equal to 1k and the
bottom resistor bypassed with a 1uf cap while I get 610mVp-p and then take
the R and C to get a total impedance, remove the cap and install my new
calculated value reistor so I can get the 610mVp-p output with JUST two
resistors.

when I calculate the bottom resistor, I get 157ohms, but that does not equal
610mVp-p on the output. The value that does give me the same amplitude is
178ohms, but I have NO idea how to calculate that value.

Thanks again
 
T

TheTechnician

Jan 1, 1970
0
Peter said:
Well that's why I'm using THIS circuit so I don't have to worry about all
the factors of the transistor.

If I can figure out how to go from a 2 resistor series circuit with the
bottom resistor bypassed by a cap and the middle of the two resistors as my
Vout to just a 2 resistor series circuit with ONLY changing the bottom value
resistor to the equivilent of the R and C combiation and get the SAME Vout,
then I will know for sure that my emitter is all set and any variations I
get in calculations will be due to temp or whatever.

This is why this circuit is important. I don't care about phase. I ONLY care
about how to go from 1kHz 4Vp-p with both resistors equal to 1k and the
bottom resistor bypassed with a 1uf cap while I get 610mVp-p and then take
the R and C to get a total impedance, remove the cap and install my new
calculated value reistor so I can get the 610mVp-p output with JUST two
resistors.

when I calculate the bottom resistor, I get 157ohms, but that does not equal
610mVp-p on the output. The value that does give me the same amplitude is
178ohms, but I have NO idea how to calculate that value.

Thanks again
What's the big problem?
E=I*R

Vr1=(4-0.610) voltage across r1
r1=1k a given
V/R=I (0.00339) find current through r1
Vr2=(0.610) a given
V/I=R find the resistance that will drop 0.61V at 3.39mA
(0.610)/(0.00339)=179.9410029499

you said it works with a 178 ohm
 
P

Peter

Jan 1, 1970
0
OK. figure out the resistance of the R and C without using any voltages
 
Peter said:
OK. figure out the resistance of the R and C without using any voltages
This has been answered before. In case you missed it:
Xc = 1/2*pi*F*C
That gives you the capacitve reactance in ohms.
If you already know the total resistance of R and C in parallel:
Rtot = (R*Xc)/(R + Xc)
Solve for R

If you know R but do not know Rtot, the equation is the same:
Rtot = (R*Xc)/(R + Xc)
Solve for Rtot

Ed
 
D

daestrom

Jan 1, 1970
0
This has been answered before. In case you missed it:
Xc = 1/2*pi*F*C
That gives you the capacitve reactance in ohms.
If you already know the total resistance of R and C in parallel:
Rtot = (R*Xc)/(R + Xc)
Solve for R

Sorry, this doesn't give the correct results. Xc is displaced 90 degrees
from R. A 10 ohm resistor and a capacitive reactance of 10 ohms in parallel
with it does *not* give you (10*10)/(10+10) = 5 ohms impedance.

With parallel components, to combine them it is useful to calculate the
individual conductance and susceptance, then add them (being sure to account
for resistance vs. reactance), then convert the resulting admittance back to
impedance (you just got to love all these terms ;-)

For example, a 10 ohm resistor has a conductance of 0.1 mho. Similarly, the
capacitor with 10 ohm reactance has a susceptance of 0.1 mho. These combine
to give the complex admittance

For parallel circuits....
admittance = sqrt (conductance^2 + susceptance^2)
= sqrt (0.1^2 + 0.1^2) = 0.1414 mho

So the impedance is 1/0.1414 = 7.07 ohm

Lumped all together, for a resistance R and capacitive reactance Xc....

Z = 1 / (sqrt( (1/R)^2 + (1/Xc)^2))

http://searchsmb.techtarget.com/sDefinition/0,,sid44_gci212333,00.html

If the resistance or reactance is significantly larger than the other, your
formula will give you *approximately* correct results. But if the ohmic
values are near each other, the proper calculation is what I just described.

daestrom
 
daestrom said:
Sorry, this doesn't give the correct results. Xc is displaced 90 degrees
from R. A 10 ohm resistor and a capacitive reactance of 10 ohms in parallel
with it does *not* give you (10*10)/(10+10) = 5 ohms impedance.
Thanks - you are correct and my answer was wrong.
Started thinking DC (obviously) when its AC. I hate
it when I do that.
Ed
 
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